Admitting the ansatz
$$ψ=e^{i(kx-ω t)} \tag{1}$$
then
$$k^2=-ψ^{-1} \frac {∂^2ψ}{∂x^2} \tag{2}$$
and
$$ω=iψ^{-1} \frac {∂ψ}{∂t} \tag{3}$$
If one admits that the total energy ($E$) is related to momentum ($p$) as $E=\frac{p^2}{2m}+U$, admiting also the De Broglie relations $E=ħω$; $p=ħk$ it follows that
$$\frac {-ħ^2}{2m} \frac {∂^2ψ}{∂x^2}+Uψ= iħ \frac {∂ψ}{∂t} \tag{4}$$
This is Schrödinger's equation. This equation is said to be non relativistic because of its use of $E= \frac{p^2}{2m}+U$ (rigorously speaking though, it is non relativistic because it is not Lorentz invariant).
However, starting from the relativistic total energy equation
$$E=\frac{1}{ \sqrt{1- \frac{v^2}{c^2}}}mc^2=T+mc^2 \tag{5}$$
Where, $T$ is the kinetic energy and $mc^2$ the particle’s self energy. Now, using the expansion of $\frac{1}{ \sqrt{1- \frac{v^2}{c^2}}}mc^2$
$$E=mc^2 + \frac{mv^2}{2} + \frac{3mv^4}{8c^2} + \frac{5mv^6}{16c^4}+…$$
and ignoring members dividing by $c$ (because we are considering $v\ll c$). It becomes
$$E=\frac{1}{2} mv^2+mc^2=T+mc^2 = \frac{p^2}{2m}+mc^2 \tag{6}$$
or
$$\frac{p^2}{2m}+mc^2=ħω=\frac{ħ^2k^2}{2m}+mc^2 \tag{7}$$
So, $mc^2$ do not vanishes even under classical approximation.
Admitting that Planck's and De Broglie's equations holds in every situation and that $E$ in Planck equation is the total energy, substituting equation (2) and (3) into (7) the Schrödinger equation “would” have the form
$$\frac {-ħ^2}{2m} \frac {∂^2ψ}{∂x^2}+mc^2 ψ = iħ \frac {∂ψ}{∂t} \tag{9}$$
Now we could postulate this equation, making the steps of getting it less fundamental then the end result.
I tried to consider that $T \ll mc^2$ in Schrödinger equation, but I realize that an electron in hydrogen atom moving with half the speed of light (using classical equations as we are analyzing Schrödinger’s equation) it would have less than $\rm 100keV$ ($\rm ≈64keV$ if my math is not wrong) of kinetic energy, but $\rm 511keV$ of self energy.
So, my question is: why Schrödinger equation do not have an $mc^2$ term, if $ħω$ is supposed to be the total energy and not just the kinetic energy.
Best Answer
Just because to introduce the constant added term $mc^2I$ to the Hamiltonian operator would be equivalent to redefine $\psi \to \psi'= e^{imc^2 t/\hbar}\psi$. This sort of phases do not matter in QM. You cannot see them by measuring any observable. Pure states are actually operators of the form $|\psi \rangle \langle \psi|$ and you see that these phases cancel each other.
Instead, if the mass were replaced by a mass operator with discrete spectrum the picture would change. In the classical limit the rapid temporal oscillations of the phases (I am assuming that the mass is big if compared with the typical energies of the system), would destroy the coherence of superpositions of different masses giving dynamically rise to superselection rule of the mass Bargmann's superselection rule (see here or here).